Question

A 40 kg stone is dropped from a height of 150m. What is the Potential Energy and Kinetic Energy
a. at the top?
b. at t=1s?
c. when it strikes the ground?

Answer 1

potential energy when the stone at 150m of height can be determined with E=mhg.
kinetic energy can be count by E=(1/2)mv².

When the stone at 150m high, it posses the maximun gravitational energy, and it is gradually convert into kinetic energy when the stone is falling to the ground. Because of this gravitational energy decrease but kinetic energy increase. And the 'initial gravitational energy' is equal to the kinetic energy when it reach the ground.

To find the kinetic energy at 1s, i guess u first need to know the velocity it possed. You can use V² = U² + 2as, U suppose to be 0 and a suppose to be gravitation acceleration.

Answer 2

the problem is i dont know what's the velocity

Answer 3

with v²=u² + 2as
u , initial velocity is zero
a , gravitational acceleration.
so: V² = (0)² + 2(10m/s²)(1s)
v=sqrt(20)m/s

Answer 4

oh. thanks!i'll try it

Answer 5

At t = 1 s, velocity is 10 m/s, not √20

Answer 6

instead of that, i used Vf=Vi+gt

Answer 7

do you know the formula of potential energy? @Yanyan

Answer 8

yes

Answer 9

mgh right

Answer 10

yep

Answer 11

m=40kg g=9.8 m/s^2 h=150,

Answer 12

P.E. in b is 0, right?

Answer 13

sp P.E=40*9.8*150=58800J

Answer 14

58800J at top=P.E and 0J=K.E

Answer 15

i'm confused in b and c.

Answer 16

b/ mg(150-1/2*g*t^2)
1/2m*v^2=1/2*m*(gt)^2
c/1/2*m*(2gh)^(1/2)